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An Introduction to the Bergman Projection and Kernel

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An Introduction to the Bergman Projection and Kernel An Introduction to the Bergman Projection and Kernel A Thesis Presented in Partial Fulfillment of the Requirements for the Degree Master of Mathematical Sciences in the Graduate School of The Ohio State University By Philip Koch, B.S. Graduate Program in the Department of Mathematics The Ohio State University 2016 Master's Examination Committee: Dr. Kenneth Koenig, Advisor Dr. Rodica Costin Dr. Jeffery McNeal c Copyright by Philip Koch 2016 Abstract In this thesis, we discuss the fundamental concepts of the Bergman projection and kernel. In particular, we prove that the Bergman kernel on a given domain Ω can be characterized in terms of an orthonormal basis of A2(Ω). This result is central to the theory of the Bergman kernel and will therefore be shown in full detail. We calculate the Bergman kernel on several domains such as the unit disk, the unit ball in C2 and Cn, and the annulus. In addition, we show that given two conformally equivalent domains Ω1; Ω2 ⊆ C, one can represent the Bergman kernel of Ω1 in terms of the Bergman kernel of Ω2. ii I dedicate this thesis to my wife, my parents, and my brothers. iii Acknowledgments I would like to thank several people who helped make this possible. First, I thank my advisor, Dr. Kenneth Koenig, for his support and guidance. His knowledge and wisdom were invaluable to my progress and growth as a mathematician and I feel extremely lucky to have worked with him. Second, I would like to thank my committee members, Dr. Jeffery McNeal and Dr. Rodica Costin. Their comments and insight were greatly appreciated. In particular, Dr. Costin's final edits were very beneficial. Third, I thank my wife, Ayla, for her love and encouragement. She was my rock throughout this process. For taking care of the dishes and soda cans that I left on the table while working, for comforting me when I needed a break, and for all the other little things you do, thank you. Fourth, I thank my brother, Thayer, for his advice and friendship. He was there when I needed someone to talk to and always available to play a video game when I wanted to relax. Last I thank my parents, Lorin and Tandy, for their love and encouragement. So much of what I am today is because of them. You both are truly wonderful people and I could not have asked for better parents. iv Vita June 2012 . B.S. Mathematics, University of Washington 2014-present . .Graduate Teaching Associate, Ohio State University. Fields of Study Major Field: Mathematics v Table of Contents Page Abstract . ii Dedication . iii Acknowledgments . iv Vita......................................... v 1. Motivation . 1 2. The Bergman Projection and Kernel . 9 3. The Bergman Kernel in C2 and Cn ..................... 28 4. The Bergman Kernel and Conformal Mappings . 36 Bibliography . 47 vi Chapter 1: Motivation We start by stating our topic of study: integral operators and their associated kernels over C (and later we will look into those over Cn). In particular we begin with the Cauchy integral formula. A list of notation used in this thesis appears at the end of this chapter. Theorem 1.1. Let Ω ⊆ C be a bounded domain with piecewise smooth boundary. If f is holomorphic on Ω and f extends smoothly to the boundary of Ω, then 1 Z f(w) f(z) = dw (1.1) 2πi @Ω w − z for any point z 2 Ω. [1, p. 113] The Cauchy integral formula is an amazing fact and the very thing that piqued this author's interest into complex analysis. We ask ourselves \what else can be said about this?" For one thing, it turns out for each f 2 C(@Ω), the Cauchy integral is holomorphic in Ω, which we'll show in Corollary 1.3 below. Theorem 1.2. Let F (z; s) be defined for (z; s) 2 Ω × [0; 1], where Ω is an open set in C. Suppose F satisfies the following properties: (i) F (z; s) is holomorphic in z for each s. (ii) F is continuous on Ω × [0; 1]. 1 Then the function defined on Ω by Z 1 f(z) = F (z; s) ds 0 is holomorphic. [3, p. 56] Corollary 1.3. Let Ω ⊆ C be a bounded open subset with smooth boundary and let f 2 C(@Ω). Then the Cauchy transform of f, 1 Z f(w) C(f)(z) = dw (1.2) 2πi @Ω w − z is holomorphic in Ω. Thus the Cauchy transform, C, can be thought of as an integral operator with 1 associated kernel C(z; w) = . 2πi(w − z) Proof: Let γ : [0; 1] ! @Ω be a smooth parameterization of @Ω and let z 2 Ω. Then, by setting w = γ(t) in (1.2), 1 Z f(w) 1 Z 1 C(f)(z) = dw = F (z; t) dt; (1.3) 2πi @Ω w − z 2πi 0 f(γ(t)) where F (z; t) = γ0(t). We claim that F is holomorphic in z for each t and γ(t) − z that F is continuous on Ω × [0; 1] in order to apply Theorem 1.2. Notice that for any t 2 [0; 1], z 6= γ(t) for all z 2 Ω. For if so, then there exists ◦ t0 2 [0; 1] such that z = γ(t0) 2 @Ω. This would imply that z 62 Ω . Since Ω is open, we would therefore have z 62 Ω. Hence we have a contradiction. And so, γ(t) − z 6= 0 for all t 2 [0; 1] and z 2 Ω. f(γ(t)) Therefore, for each t 2 [0; 1], F (z; t) = γ0(t) is holomorphic in z. Further, γ(t) − z as f 2 C(@Ω), F (z; t) is continuous on Ω × [0; 1]. Thus, by Theorem 1.2, C(f)(z) is holomorphic. 2 With all the wonderful results that complex analysis holds we have to ask ourselves several questions. Was the Cauchy integral formula a fluke or are there more equations like this? If so, what properties do they possess? In addition, are they generalizable? We plan to address each of these questions. In the coming chapters we will discuss the latter two, but for now we draw our attention to the first. Before doing so we recall two classic results in complex analysis mentioned below. Theorem 1.4. Let f be a holomorphic function on an open set Ω ⊆ C. Then for any zo 2 Ω and 0 < r < dist(z0;@Ω) 1 Z 2π 1 Z f(z ) = f(z + reiθ) dθ = f(z) dV (z): (1.4) 0 2π 0 πr2 0 Dr(z0) The first equality of (1.4) is the mean value property for circles and the second is the mean value property for disks. We will utilize the latter in the proof of an important result in chapter 2. Whereas, for our purposes, the former will only be used in obtaining the property for the disk. Proof: Let f be a holomorphic function on an open set Ω ⊆ C and let z0 2 Ω. Then for any 0 < r < dist(z0;@Ω), Dr(z0) ⊆ Ω. Hence, by the Cauchy integral formula, 1 Z f(w) f(z ) = dw: (1.5) 0 2πi w − z @Dr(z0) 0 iθ iθ By the substitution w = z0 + re for 0 ≤ θ ≤ 2π, dw = ire and 1 Z 2π f(z + reiθ) 1 Z 2π 0 iθ iθ (1.6) f(z0) = iθ ire dθ = f(z0 + re ) dθ 2πi 0 re 2π 0 3 iθ thereby establishing the first equality. Next, by setting z = z0 + se for 0 ≤ s ≤ r and 0 ≤ θ ≤ 2π we have 1 Z 1 Z r Z 2π f(z) dV (z) = f(z + seiθ)s dθds 2π 2π 0 Dr(z0) 0 0 Z r = f(z0)s ds (1.7) 0 r2 = f(z ) : 0 2 Note that the middle equality was due to the mean value property for circles es- tablished in (1.6). Therefore, by dividing by r2=2, we obtain the second equality. The answer to the first question is that \no, this wasn't some fluke." There are many different equations like the Cauchy integral formula. In particular, we will explore one in detail but do note that there are others to be studied as well. Theorem 1.5 (The Bergman kernel on the unit disk). For all f 2 A2(D) and z 2 D, 1 Z f(w) f(z) = 2 dV (w): (1.8) π D (1 − zw) Notice that the integral defined above is a convergent integral. Indeed, as jzj < 1 and jwj < 1, jzwj = jzjjwj < jzj and 1 − jzj > 0. Then, by the reverse triangle inequality, j1 − zwj ≥ 1 − jzwj > 1 − jzj. Thus, Z 2 Z 2 1 1 2 dV (w) ≤ 2 dV (w) < 1; (1.9) D (1 − zw) D (1 − jzj) and we have the convergence of (1.8) by means of the Cauchy-Schwarz inequality. Proof: Let f 2 A2(D) and fix z 2 D. By setting w = reiθ with 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π, we have dV (w) = dxdy = rdrdθ. Hence, Z f(w) Z 1 Z 2π f(reiθ) (1.10) 2 dV (w) = −iθ 2 r dθdr: D (1 − zw) 0 0 (1 − zre ) 4 −idw Since w = reiθ, dw = ireiθdθ = iwdθ. Therefore dθ = , e−iθ = r , and w w Z 1 Z 2π f(reiθ) Z 1 Z f(w) −idw r dθdr = r dr (1 − zre−iθ)2 r2 2 w 0 0 0 @D (1 − z w ) Z 1 Z −irf(w) dw = dr (1.11) w−zr2 2 w 0 @D ( w ) Z 1 Z −irwf(w) = 2 2 dwdr: 0 @D (w − zr ) .
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